A. (140x110)+[(50x70)/2] = 15,400+1750= 17,150cm^2
b. (4x6)/2= 12cm^2
Answer:
2X/(3+X)
Step-by-step explanation:
X = 3m/2-m
Cross multiply both sides
X × (2-m)= 3m
2X-Xm= 3m
2X= 3m+Xm
2X= m(3+X)
Divide both sides by the coefficient of m which is (3+X)
m= 2X/(3+X)
False i think
because 11x11=121
and the part where there is missing its 5x2 which is 10
121-10 is 111
so no i dont think it’s approximately 101cm
If the triangles above are congruent then the length of EF should be the same as the length of BC which is 33.
Hope this helps!
Answer:
(1-cos2A) /(1+cos2A) =tan²A
Proof:
We know that,
cos(A+B) =cosA.cosB-sinA.sinB
=>cos2A=cos(A+A)
=>cos2A=cosA.cosA - sinA.sinA
=>cos2A=cos²A-sin²A
=>cos2A=(cos²A-sin²A)/(cos²A+sin²A
Since {cos²A+sin²A=1}
Divide the numerator & the denominator by (cos²A) to get,
cos2A = {(cos²A-sin²A) ÷cos²A} / {(cos²A+sin²A) ÷cos²A}
cos2A ={(1-tan²A)/(1+tan²A)}
Then,
1-cos2A = 1-[{(1–tan²A)/(1+tan²A)}]
1-cos2A =(1+tan²A-1+tan²A)/(1+tan²A)
1-cos2A=(2tan²A)/(1+tan²A)
And now.......
1+cos2A=1+[{(1-tan²A)/(1+tan²A)}]
1+cos2A={1+tan²A+1-tan²A}/{1+tan²A}
1+cos2A=2/(1+tan²A)
So now,
(1-cos2A)/(1+cos2A)= {2tan²A/(1+tan²A)}÷{2/(1+tan²A)}
={(2tan²A)(1+tan²A)}÷{2(1+tan²A)}
=tan²A
Step-by-step explanation:
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